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A Game of Tennis (Posted on 2007-07-05) Difficulty: 3 of 5
When Justine serves a game of tennis against Maria, Justine wins any given point 80% of the time.
  • If at some point, the game is at deuce (40-40), what is the probability Justine will eventually win?
  • If it's advantage Justine ("ad in" in tennis lingo--functionally equivalent to 40-30), what are her chances?
  • What is her probability of winning if the advantage is to her opponent, Maria, (equivalent to 30-40)?

Scoring in tennis is, 0, 15 (one successful point), 30 (two), 40 (three). The next point is a win unless the opponent already has 40 as well, as you must win by two points. The same player serves for the entire game, so each point within one game has the same probability given. Then the serve switches for the next game in the set or match.

  • What is Justine's probability of winning the game, calculated when the score is still 0-0?
  • What are the probabilities from any given state of the game?

What about when Maria is serving, if she wins any given point against Justine 75% of the time?

  Submitted by Charlie    
Rating: 3.5000 (4 votes)
Solution: (Hide)
Call the probability of Justine's making her point p (equal to 0.80 in this case), and that her opponent will take the point as q (0.20 in this case).

Call the probability of Justine's winning the game from deuce d. And from ad iN (advantage Justine), n, and from ad ouT, t.

The probability is p that deuce will be converted to ad in, and q that it will be converted to ad out. Likewise the probability is p that ad in would be converted to a win, and q that it would be converted to deuce. Finally it is probability p that ad out would be converted to deuce and q that it would be converted to a loss. So:

d = pn + qt
n = p + qd
t = pd

so

d = p(p+qd) + qpd
d = p^2 + pqd + pqd
d - 2pqd = p^2
d(1-2pq) = p^2
d = p^2 / (1-2pq)

Once you have d, you can find n and t from the remaining formulae above.

In the case of p = 0.80, or 8/10, the probability of a win from deuce comes out to 16/17 or about .941176; from ad in, 84/85 or about .988235; and from ad out, 64/85 or about .752941.

Probabilities from lower given scores can be found by proceding backwards. For example, p(40-15) = p + q*p(40-30) = p + qd. (The hyphens indicate the separator of the scores, not subtraction). Or p(30-15) = p*p(40-15) + q*p(30-30). (Where p(30-30) is of course p*p(40-30) + q*p(30-40).)

Going backward in this manner, the probability, calculated at the beginning of the game, that Justine wins is 0.978221.

A set of tables of probabilities follows. Each table is headed by the percent of points that are expected to be scored by the server, and shows the probability of the server's winning the game, given any state of the game (score thus far) that might occur in the game.

    p 50%             60%             70%             80%             90%
from
 0- 0 0.500000   0- 0 0.735729   0- 0 0.900789   0- 0 0.978221   0- 0 0.998552
 0-15 0.343750   0-15 0.576222   0-15 0.789018   0-15 0.930033   0-15 0.990551
 0-30 0.187500   0-30 0.368862   0-30 0.587831   0-30 0.795106   0-30 0.944144
 0-40 0.062500   0-40 0.149538   0-40 0.289776   0-40 0.481882   0-40 0.720110
15- 0 0.656250  15- 0 0.842068  15- 0 0.948691  15- 0 0.990268  15- 0 0.999441
15-15 0.500000  15-15 0.714462  15-15 0.875241  15-15 0.963765  15-15 0.995707
15-30 0.312500  15-30 0.515077  15-30 0.715569  15-30 0.873412  15-30 0.969037
15-40 0.125000  15-40 0.249231  15-40 0.413966  15-40 0.602353  15-40 0.800122
30- 0 0.812500  30- 0 0.927138  30- 0 0.980169  30- 0 0.996894  30- 0 0.999856
30-15 0.687500  30-15 0.847385  30-15 0.943672  30-15 0.986353  30-15 0.998671
30-30 0.500000  30-30 0.692308  30-30 0.844828  30-30 0.941176  30-30 0.987805
30-40 0.250000  30-40 0.415385  30-40 0.591379  30-40 0.752941  30-40 0.889024
40- 0 0.937500  40- 0 0.980308  40- 0 0.995810  40- 0 0.999529  40- 0 0.999988
40-15 0.875000  40-15 0.950769  40-15 0.986034  40-15 0.997647  40-15 0.999878
40-30 0.750000  40-30 0.876923  40-30 0.953448  40-30 0.988235  40-30 0.998780
40-40 0.500000  40-40 0.692308  40-40 0.844828  40-40 0.941176  40-40 0.987805

   p  55%             65%             75%             85%             95%
from
 0- 0 0.623149   0- 0 0.829645   0- 0 0.949219   0- 0 0.992863   0- 0 0.999908
 0-15 0.458030   0-15 0.689214   0-15 0.870117   0-15 0.969215   0-15 0.998783
 0-30 0.270895   0-30 0.476563   0-30 0.696094   0-30 0.879353   0-30 0.985508
 0-40 0.099660   0-40 0.212897   0-40 0.379688   0-40 0.595578   0-40 0.855007
15- 0 0.758246  15- 0 0.905261  15- 0 0.975586  15- 0 0.997036  15- 0 0.999967
15-15 0.611140  15-15 0.803719  15-15 0.928125  15-15 0.985073  15-15 0.999481
15-30 0.410996  15-30 0.618536  15-30 0.801563  15-30 0.929431  15-30 0.992376
15-40 0.181200  15-40 0.327534  15-40 0.506250  15-40 0.700680  15-40 0.900007
30- 0 0.878605  30- 0 0.959937  30- 0 0.991406  30- 0 0.999147  30- 0 0.999992
30-15 0.774894  30-15 0.903433  30-15 0.970313  30-15 0.994892  30-15 0.999855
30-30 0.599010  30-30 0.775229  30-30 0.900000  30-30 0.969799  30-30 0.997238
30-40 0.329455  30-40 0.503899  30-40 0.675000  30-40 0.824329  30-40 0.947376
40- 0 0.963460  40- 0 0.990363  40- 0 0.998438  40- 0 0.999898  40- 0 1.000000
40-15 0.918800  40-15 0.972466  40-15 0.993750  40-15 0.999320  40-15 0.999993
40-30 0.819554  40-30 0.921330  40-30 0.975000  40-30 0.995470  40-30 0.999862
40-40 0.599010  40-40 0.775229  40-40 0.900000  40-40 0.969799  40-40 0.997238

So for Maria serving Justine, Maria's chance of winning the game from deuce would be 90%; from ad in or 40-30, 97.5% (or 39/40) and from ad out or 30-40, 67.5% (or 27/40). At the beginning of the game, you'd say 94.9219...%.

In rational numbers, the relevant figures are:

80 %
p = 4/5; q = 1/5
from                          prob of win
40-40  (deuce)                 16/17
40-30                          84/85
40-15  p + q*p(40-30)          424/425
40-0   p + q*p(40-15)          2124/2125
30-40                          64/85
30-30  p*p(40-30) + q*p(30-40) 16/17
30-15  p*p(40-15) + q*p(30-30) 2096/2125
30-0   p*p(40-0) + q*p(30-15)  10592/10625
15-40  p*p(30-40)              256/425
15-30  p*p(30-30) + q*p(15-40) 1856/2125
15-15  p*p(30-15) + q*p(15-30) 2048/2125
15-0   p*p(30-0) + q*p(15-15)  52608/53125
0-40   p*p(15-40)              1024/2125
0-30   p*p(15-30) + q*p(0-40)  8448/10625
0-15   p*p(15-15) + q*p(0-30)  49408/53125
0-0    p*p(15-0) + q*p(0-15)   51968/53125

75 %
p = 3/4; q = 1/4
from                          prob of win
40-40  (deuce)                 9/10
40-30                          39/40
40-15  p + q*p(40-30)          159/160
40-0   p + q*p(40-15)          639/640
30-40                          27/40
30-30  p*p(40-30) + q*p(30-40) 9/10
30-15  p*p(40-15) + q*p(30-30) 621/640
30-0   p*p(40-0) + q*p(30-15)  1269/1280
15-40  p*p(30-40)              81/160
15-30  p*p(30-30) + q*p(15-40) 513/640
15-15  p*p(30-15) + q*p(15-30) 297/320
15-0   p*p(30-0) + q*p(15-15)  999/1024
0-40   p*p(15-40)              243/640
0-30   p*p(15-30) + q*p(0-40)  891/1280
0-15   p*p(15-15) + q*p(0-30)  891/1024
0-0    p*p(15-0) + q*p(0-15)   243/256

This puzzle is not based on any actual statistics of any real people named Justine or Maria. It was inspired by a supposedly counterintuitive result in Julian Havil's Book, Nonplussed: Mathematical Proof of Implausible Ideas, that when a player has a probability of winning the point on her service greater than 0.919643... (p > 0.919643..., the root of the equation 8p^3 - 4p^2 - 2p - 1 = 0), then p(0-0) > p(40-30). But when p > 0.919643, the value of p(0-0) is over .999391, which means that her service would be broken only once in at least 1642 games; the players are clearly not well matched, and it would be amazing if it got to 40-30; it would be hard enough for the lesser player to get to merely 40-15. Mr. Havil appears to believe that such high percentages would apply among evenly matched players, but that's not the case: the server advantage is not as great as he thinks it is.

DECLARE FUNCTION pWin# (a#, b#)
DEFDBL A-Z
CLEAR , , 9999
DIM score(4)
DIM SHARED p, q, d, t, n

score(1) = 15
score(2) = 30
score(3) = 40
score(4) = 99

CLS

FOR pp = 50 TO 90 STEP 10
  p = pp / 100
  q = 1 - p

  d = p * p / (1 - 2 * p * q)
  t = d * p
  n = p + q * d

  col = (pp / 10 - 5) * 15 + 1
  LOCATE 2, col + 6: PRINT USING "##&"; pp; "%";
  ln = 2
  FOR own = 0 TO 3
   FOR opp = 0 TO 3
    LOCATE 2 + ln, col
    PRINT USING "##&## #.######"; score(own); "-"; score(opp); pWin(own, opp)
    ln = ln + 1
   NEXT
  NEXT
NEXT


FOR pp = 55 TO 95 STEP 10
  p = pp / 100
  q = 1 - p

  d = p * p / (1 - 2 * p * q)
  t = d * p
  n = p + q * d

  col = (pp / 10 - 5.5) * 15 + 1
  LOCATE 21, col + 6: PRINT USING "##&"; pp; "%";
  ln = 21
  FOR own = 0 TO 3
   FOR opp = 0 TO 3
    LOCATE 2 + ln, col
    PRINT USING "##&## #.######"; score(own); "-"; score(opp); pWin(own, opp)
    ln = ln + 1
   NEXT
  NEXT
NEXT

FUNCTION pWin (a, b)
  IF a = 3 AND b = 3 THEN
   pWin = d
  ELSEIF a = 3 AND b = 2 THEN
   pWin = n
  ELSEIF a = 2 AND b = 3 THEN
   pWin = t
  ELSEIF a = 4 THEN
   pWin = 1
  ELSEIF b = 4 THEN
   pWin = 0
  ELSE
   pWin = p * pWin(a + 1, b) + q * pWin(a, b + 1)
  END IF
END FUNCTION

Comments: ( You must be logged in to post comments.)
  Subject Author Date
Line JudgeSteve Herman2007-07-07 07:53:07
Solution first questionsClaire2007-07-06 05:50:31
re(3): First question - disagreeCharlie2007-07-06 01:11:52
Some Thoughtsre(2): First question - disagreeOld Original Oskar!2007-07-05 22:33:35
re: First questionVernon Lewis2007-07-05 15:19:48
Some ThoughtsFirst questionFederico Kereki2007-07-05 14:05:12
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